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The question is in the title.

Equation $\sum_{i=1}^n x_i^3 = 0$ has no non-trivial integer solutions for $n=3$. For $n=4$, there are known descriptions of all integer/rational solutions, see

Choudhry, Ajai. "On equal sums of cubes." The Rocky Mountain journal of mathematics (1998): 1251-1257.

My question is about the next case, $n=5$. Because the equation is homogeneous, integer and rational solutions are related in an obvious way. Some families of solutions are known, see for example this link and this one, but the question is to describe ALL solutions.

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  • $\begingroup$ Are you happy with my answer or is there something you would like clarified? $\endgroup$ Commented Jul 21, 2023 at 9:49
  • $\begingroup$ Yes, I am happy, thank you for the answer. There are two answers, one that the solution is impossible, another one that it might be possible, so I was not sure which answer to accept. I have now accepted your answer. $\endgroup$ Commented Jul 22, 2023 at 11:25

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This is not possible in any meaningful way.

In fact the variety you describe defines a smooth cubic threefold $X$ in $\mathbb{P}^4$. By a famous theorem of Clemens and Griffiths these are not even rational varieties over $\mathbb{C}$. This means that these are no way to parametrise the $\mathbb{C}$-points, let alone the $\mathbb{Q}$-points.

Here are some partial attempts which fail.

Firstly this variety $X$ is unirational over $\mathbb{Q}$ which means that it admits a dominant rational map $\mathbb{P}^3 \dashrightarrow X$ over $\mathbb{Q}$. This gives a Zariski dense set of rational points on $X$. But it is impossible to write all rational points this way, even allowing finitely many maps. This follows from the fact that $X(\mathbb{Q})$ is not thin (see Theorem 3.1 of [1]).

Next one expects that $X$ satisfies weak approximation. This means that $X(\mathbb{Q})$ should be dense in $X(\mathbb{R})$ and $X(\mathbb{Q}_p)$ for all primes $p$. So one can ``describe'' rational points by stipulating that you have rational points close to some given collection of real and $p$-adic points. But this is a partial solution, and in any case no one is able to prove that weak approximation holds with current tools.

This is nothing special about your variety $X$. All these properties are expected to hold for any rationally connected variety with a rational point which is not a rational variety. (In general one needs to replace weak approximation by weak weak approximation.)

[1] Demeio, Julian Lawrence. Elliptic fibrations and the Hilbert property. Int. Math. Res. Not. IMRN 2021, no. 13, 10260--10277.

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    $\begingroup$ I believe it is still unknown whether cubic threefolds are stably rational (which would have many of the same consequences as rationality for the problem of “parameterizing” rational points). $\endgroup$ Commented Jul 16, 2023 at 0:40
  • $\begingroup$ Probably something else was meant when you wrote “replace weak approximation with weak approximation” $\endgroup$ Commented Jul 16, 2023 at 0:42
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    $\begingroup$ @Stanley, that's not what Daniel wrote. $\endgroup$ Commented Jul 16, 2023 at 12:20
  • $\begingroup$ Indeed I wrote "weak weak approximation" (this is not a typo!). You can read about this and related background on thin sets here: en.wikipedia.org/wiki/Thin_set_(Serre) $\endgroup$ Commented Jul 17, 2023 at 9:04
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    $\begingroup$ I did indeed fail reading comprehension and I apologize! I will leave the comment here so everyone can have a chuckle at my expense. $\endgroup$ Commented Jul 17, 2023 at 11:38
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I am just posting my comment as another, rather contradictory answer. It might indeed be possible both to parameterize rational points and even prove weak approximation in an easy, geometric way, in spite of the theorem of Clemens-Griffiths that all smooth cubic threefolds are irrational. This is because the more relevant geometric property is "stable rationality" (or perhaps something even weaker). Weak approximation holds for all stably rational varieties. So far nobody has been able to either prove or disprove stable rationality of cubic threefolds (the technique of Voisin using "decomposition of the diagonal" has so far been unable to resolve this problem).

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    $\begingroup$ Thanks for this very nice observation which I had overlooked! $\endgroup$ Commented Jul 17, 2023 at 9:05

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